Assume that \(X\cong PSp_{n}(q)\)  with \(n \geqslant 4\) . Then \(X_{x}\)  lies in a large maximal
geometric subgroup \(Y\)  of \(X\)  by Theorem REF . Hence, \(Y\)  is
one of the groups listed in [1]}, Proposition 4.22. It follows from
Theorem 3.5 of [2]} and Proposition 5.2.15.(i) of [3]}, that \(Y\notin \mathcal {C}_{1}(X)\)  and \(Y\notin \mathcal {C}_{2}(X)\cup \mathcal {C}_{5}(X)\) , respectively, since \(\left( \Phi _{nf}^{\ast }(p),\left|Y\right|\right) >1\) . For the same reason \((X,Y)\ne (PSp_{8}(3),2^{6}.\Omega _{6}^{-}(2))\) , \((PSp_{4}(5),2^{4}.\Omega _{4}^{-}(2))\) . Thus, one of the following holds:

\(Y\)  is a \(\mathcal {C}_{8}\) -subgroup of \(X\) ;

\(Y\)  is a \(\mathcal {C}_{3}\) -subgroup of \(X\)  of type \(Sp_{n/2}(q^{2})\) , \(Sp_{n/3}(q^{3})\)  or \(GU_{n/2}(q)\) ;

\((X,Y)\)  is either \((PSp_{4}(7),2^{4}.O_{4}^{-}(2))\)  or \((PSp_{4}(3),2^{4}.\Omega _{4}^{-}(2))\) .

Assume that Case (1) holds. Then \(Y\cong O_{n}^{-}(q)\)  with \(q\)  even by [3]}, Proposition 4.8.6.(II), as \(\left( \Phi _{nf}^{\ast }(p),\left|Y\right|\right) >1\) . Also \(X_{x}=Y\)  by Lemma REF (1) and by [3]}, Proposition 4.8.6.(I). Then \(k^{2}=q^{n/2}(q^{n/2}-1)\) , since \(k^{2}=\left[ X:X_{x}\right] \) , and hence \(q^{n/2}-1\)  is a square.
However, this is impossible by [6]}, A3.1, since \(n\geqslant 4\) .
Assume that Case (2) holds. If \(Y\)  is a \(\mathcal {C}_{3}\) -subgroup of \(X\)  of
type \(Sp_{2i}(q^{t})\) , \(t=2,3\) , where \(n=2it\) , then \(Y\cong PSp_{2i}(q^{t})\cdot Z_{t}\)  by [3]}, Proposition 4.3.10.(II). Also \(X_{x}=Y\)  by Lemma REF (1) and by [3]}, Proposition 4.3.10.(I). It
follows from Lemma REF (1) that \(\left|X\right|\leqslant \left|\mathrm {Out(}X\mathrm {)}\right|^{2}\cdot \left|X_{x}\right|\cdot \left|X_{x}\right|_{p^{\prime }}^{2}\) ,
which in turn yields the first (centered) inequality at page 21 of [9]} (see part (4) in the proof of Proposition 4.4). Then \(i=1\)  for \(t=3\) 
and \(i\leqslant 2\)  for \(t=2\) . The case \(t=3\)  is clear, whereas some remarks
should be done for \(t=2\) . In [9]} the authors assert that the
replication number of the 2-designs they investigate divides \(\left|\mathrm {Out(}X\mathrm {)}\right|(q^{n/t}-1)\)  by making use of
their their Lemmas 3.6 and 3.7.(c)–(d). In our context, Lemmas 3.6 of [9]} still works. Indeed, it is a general group-theoretical result. The role
of Lemma 3.7(c) is replaced by the inequality \(\left( r/\lambda \right)^{2}>v\) , since \(r=(k+1)\lambda \)  and \(v=k^{2}\) . Finally, the role of Lemma
3.7.(d) is replaced by Lemma REF (3). Thus, we get \(r/\lambda \mid \left|\mathrm {Out(}X\mathrm {)}\right|(q^{4i}-1)\) . At this point we
may proceed as in [9]} to gain \(q^{4i^{2}/8}<k^{2}<\left( r/\lambda \right) ^{2}<2q^{8i+1}\) , as \(\left|\mathrm {Out(}X\mathrm {)}\right|^{2}\leqslant 2q\) . Therefore \(i\leqslant 2\) .
Assume that \(t=3\)  and \(i=1\) . Then \(X\cong PSp_{6}(q)\)  and \(X_{x}\cong PSp_{2}(q^{3})\cdot Z_{3}\) . Hence, \(k^{2}=q^{3}(q^{2}-1)^{2}(q^{2}+1)/3\) .
Since 3 does not divide \(q^{2}+1\) , this one is a square. However, this is
impossible by [6]}, A3.1.
Assume that \(t=2\)  and \(i=2\) . Then \(X\cong PSp_{8}(q)\)  and \(X_{x}\cong PSp_{4}(q^{2})\cdot Z_{2}\) . Then \(k^{2}=q^{8}\left( q^{2}-1\right)^{2}\left( q^{4}+q^{2}+1\right) /2\) , but this is impossible by [6]},
A7.1.
Assume that \(t=2\)  and \(i=1\) . Then \(X\cong PSp_{4}(q)\)  and \(X_{x}\cong PSp_{2}(q^{2})\cdot Z_{2}\) . Then \(k^{2}=q^{2}(q^{2}-1)/2\) . Then \(q\)  is odd
by [6]}, A3.1. Then \(\frac{q\pm 1}{2}=z_{1}^{2}\)  and \(2(q\mp 1)=z_{2}^{2}\)  for some positive integers \(z_{1},z_{2} \) , according to whether \(q\)  is equivalent to 1 or 3 modulo 4 respectively. So \(z_{1}^{2}\mp 1=(z_{2}/2)^{2}\) , which has no solutions again by [6]}, A3.1.
Finally, assume that \(Y\)  is a \(\mathcal {C}_{3}\) -group of type \(GU_{n/2}(q)\) .
Then \(X_{x}=Y\cong Z_{\frac{q-1}{2}}.PGU_{n/2}(q).Z_{2}\)  by Lemma REF (1) and by [3]}, Proposition 4.3.7.(I)–(II). Moreover, \(n/2\)  is odd, since \(\left( \Phi _{nf}^{\ast }(p),\left|Y\right|\right) >1\) , and hence \(n\geqslant 6\) .
Then
\(k^{2}=\frac{1}{(2,q-1)}q^{\frac{1}{8}n\left( n+2\right)}\prod _{i=1}^{n/2}(q^{i}+(-1)^{i}) \) 
and hence \(q^{n/2-1}+1\)  divides \(k^{2}\) . Moreover, (REF )
is not fulfilled for \((n/2-1,q)=(6,2)\) . Then \(\Phi _{(n-2)f}^{\ast }(p)\mid k^{2}\) , where \(\Phi _{(n-2)f}^{\ast }(p)>1\) , by [6]}, P1.7, and hence \(\Phi _{(n-2)f}^{\ast }(p)^{1/2}\mid \left|G_{B}\right|\) . Moreover,
\(\left( \Phi _{(n-2)f}^{\ast }(p),\left|X_{x}\right|\right) =1\) ,
since \(n/2\)  is odd and \(n\geqslant 6\) , and \(\left( \Phi _{(n-2)f}^{\ast }(p),\left|\mathrm {Out(}X\mathrm {)}\right|\right) =1\) , as \(\Phi _{(n-2)f}^{\ast }(p)\equiv 1 \pmod {(n-2)f}\)  by [3]}, Proposition
5.2.1.5.(i). Thus \(\left( \Phi _{(n-2)f}^{\ast }(p),\left|G_{x}\right|\right) =1\)  and \(\left( \Phi _{(n-2)f}^{\ast }(p),\left[G_{B}:X_{B}\right] \right) =1\)  as both \(\left[ G_{x}:X_{x}\right] \)  and \(\left[ G_{B}:X_{B}\right] \)  divide \(\left|\mathrm {Out(}X\mathrm {)}\right|\) . Then \(\Phi _{(n-2)f}^{\ast }(p)^{1/2}\mid \left|X_{B}\right|\)  and \(\Phi _{(n-2)f}^{\ast }(p)^{1/2}\mid \left[ X:X_{B}\right] \)  as \(\left|G_{x}\right|=\lambda (k+1)\left|G_{x,B}\right|\)  and \(\left|G_{B}\right|= k \left|G_{x,B}\right|\) . Therefore \(\Phi _{(n-2)f}^{\ast }(p)^{1/2}\)  divides
both \(\left|X_{B}^{^}\right|\)  and \(\left[ SL_{n}: X_{B}^{^}\right] \) . Moreover, \(p\)  divides \(\left|X_{B}^{^}\right|\)  by Corollary REF (ii), as \(\left|X_{B}^{^}\right|_{p}=\left|X_{B}\right|_{p}\)  and \(n\geqslant 6\) . Then \(X_{B}\)  lies in a parabolic subgroup of type \(P_{2}\)  or \(P_{1,n-1}\)  of \(SL_{n}(q)\)  by Proposition REF (3).
Assume that \(X_{B}^{^}\)  lies in a parabolic subgroup of type \(P_{2} \)  of \(SL_{n}(q)\) . Then \(X_{B}\)  fixes a line \(\ell \)  of \(PG_{n-1}(q)\) ,
and hence \(\ell \)  is non-degenerate by [3]}, Propositions 4.1.3.(II)
and 4.1.19.(II), since \(\Phi _{(n-2)f}^{\ast }(p)^{1/2}\mid \left|X_{B}^{^}\right|\) . Also \(X_{B}\leqslant SL_{2}(q)\circ Sp_{n-2}(q)\) . Since \(\Phi _{(n-2)f}^{\ast }(p)^{1/2}\)  divides \(\left|\left( X_{B}\right) _{\ell ^{\perp }}^{\ell ^{\perp }}\right|\)  and \(\left[ SL_{n-2}(q):\left(X_{B}\right) _{\ell ^{\perp }}^{\ell ^{\perp }}\right] \) , where \(\left( X_{B}\right) _{\ell ^{\perp }}^{\ell ^{\perp }}\)  denotes the group induced by \(\left( X_{B}\right) _{\ell ^{\perp }}\)  on \(\ell ^{\perp }\) , it follows that
either \(n-2=4\)  and \(q=7\) , or \(\left( X_{B}\right) _{\ell ^{\perp }}^{\ell ^{\perp }}\leqslant GL_{1}(q^{n-2})\cdot Z_{n-2}\)  and \(p\mid n-2\)  by Proposition REF (1). The former does not fulfill (REF ), the latter
implies \(\left|X_{B}\right|_{p}\leqslant q\cdot (n-2)_{p}\) . Then \(q^{\frac{n^{2}-12}{12}}\leqslant q(n-2)\)  by Corollary REF (ii), which yields \(n=6\)  and \(q \leqslant 5\) , as \(n\geqslant 6\) . However, none of these values fulfills (REF ).
Assume that \(X_{B}\)  lies in a parabolic subgroup of type \(P_{1,n-1}\)  of \(SL_{n}(q)\) . Then \(X_{B}\)  is a subgroup of \(X\)  fixing a point \(\left\langle w\right\rangle \)  of \(PG_{n-1}(q)\) . Then \(X_{B}\leqslant [q^{n-1}]:Z_{q-1}\circ Sp_{n-2}(q)\)  by [3]}, Proposition 4.1.19.(II). A
similar argument to that used above shows that \(\Phi _{(n-2)f}^{\ast }(p)^{1/2}\)  divides both \(\left|\left( X_{B}\right) _{\left\langle w\right\rangle ^{\perp }/\left\langle w\right\rangle }^{\left\langle w\right\rangle ^{\perp }/\left\langle w\right\rangle }\right|\)  and \(\left[ Sp_{n-2}(q):\left( X_{B}\right) _{\left\langle w\right\rangle ^{\perp }/\left\langle w\right\rangle }^{\left\langle w\right\rangle ^{\perp }/\left\langle w\right\rangle }\right] \) . Thus \(\left( X_{B}\right)_{\left\langle w\right\rangle ^{\perp }/\left\langle w\right\rangle }^{\left\langle w\right\rangle ^{\perp }/\left\langle w\right\rangle }\leqslant GL_{1}(q^{n-2})\cdot Z_{n-2}\)  and hence \(\left|X_{B}\right|_{p}\leqslant q^{n-1}\cdot (n-2)_{p}\) . Then \(q^{\frac{n^{2}-12}{12}}\leqslant q^{n-1}(n-2)\)  by Corollary REF (ii). Thus either \(n=6\)  or \(n=10\) , as \(n/2\)  is odd.
If \(n=6\) , then (REF ) becomes
\(k^{2}=\frac{q^{6}\left( q^{2}+1\right) \left( q-1\right) ^{2}}{(2,q-1)}\left( q^{2}+q+1\right),\) 
and we reach a contradiction by [6]}, A7.1.
If \(n=10\) , then
\(k^{2}=\frac{q^{15}\left( q-1\right) \left( q^{2}+1\right) \left( q^{3}-1\right) \left( q^{4}+1\right) \left( q^{5}-1\right)}{(2,q-1)},\) 
and we obtain a contradiction, as \(k+1\)  divides \((q+1)^{3}(q^{3}+1)(2,q-1)f\)  by Lemma REF (3).
Finally, assume that case (3) holds. If \((X,Y)=(PSp_{4}(7),2^{4}.O_{4}^{-}(2))\) , then \(\Phi _{4}^{\ast }(7)=25\)  is
the highest power of 5 dividing \(\left|X\right|\) . Since \(5\mid \left|X_{x}\right|\)  and \(\left[ X:X_{x}\right] \)  is a square, it
follows that \(5^{2}\)  divides \(\left|X_{x}\right|\)  and hence \(\left|Y\right|\) , a contradiction.
If \((X,Y)=(PSp_{4}(3),2^{4}.\Omega _{4}^{-}(2))\) , then \(5\mid \left|X_{x}\right|\)  and hence \(k\mid 2^{3}\cdot 3^{2}\) . Then \(k=3,4,8,9\) ,
since \(k+1\mid \left|G\right|\) , with \(G\leqslant PSp_{4}(3).Z_{2}\) , by Lemma REF (3). However, \(X\)  does not have \(k^{2}\)  as a transitive
permutation representation degree for any such values of \(k\) 
by [23]}.
